If the area of an equilateral triangle is increasing at the rate of 4…

The correct answer is Option (4) → 6 cm/ sec

Let the side be x cm. The area A and the perimeter P are

    \(\displaystyle A = \frac{\sqrt{3}}{4}x^2,\qquad P = 3x\)

Differentiating A with respect to t (rate of change):

    \(\displaystyle \frac{dA}{dt} = \frac{\sqrt{3}}{2}x\frac{dx}{dt}\)

Putting \(\frac{dA}{dt} = 4\sqrt{3}\) and x = 4:

    \(\displaystyle 4\sqrt{3} = \frac{\sqrt{3}}{2}(4)\frac{dx}{dt} = 2\sqrt{3}\,\frac{dx}{dt}\)

    \(\displaystyle \frac{dx}{dt} = 2\) cm/sec

Hence

    \(\displaystyle \frac{dP}{dt} = 3\frac{dx}{dt} = 6\) cm/sec